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I'm trying to think of an intuitive reasoning for why $F^{\mu\nu}F_{\mu\nu}$ and $\tilde{F}_{\mu\nu}F^{\mu\nu}$ are Lorentz invariant. By this I mean that I don't simply want to show that they remain unchanged after actually performing a Lorentz transformation and seeing that I end up with the same expressions, but some sort of 'deeper' understand of why this is so. I just can't really think of why these expressions (written out in vectors like $E^2 - B^2$ and $B \cdot E$ with some constants) would be the same for every inertial observer, while for a space-time interval I can sort of grasp this.

Is there perhaps a good reference someone could point me to?

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    $\begingroup$ You could look up the Lorentz transformations of the $B$ and $E$ fields (12.108 in Griffiths' Intro to Electrodynamics, there's a derivation there too). Then, section 12.3.3 explains how to explicitly construct the EM field tensor and its dual. Finally, exercises 12.46 and 12.50 help develop an intuition. $\endgroup$
    – Danu
    Commented Mar 12, 2014 at 14:13
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    $\begingroup$ I don't know if you this is "intuitive" for you, but if you accept that F is a tensor, then obviously its contractions are scalars. $\endgroup$
    – Enucatl
    Commented Mar 12, 2014 at 14:41
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    $\begingroup$ @user129412 Yes, the 4-vector equivalent of a dot product ($x_\mu y^\mu$) in Minkowski space is a scalar, just as the Euclidean dot product in $\mathbb{R}^3$ ($x_iy^i$). The fact that we can arrange non-invariant numbers into 4-vectors with an invariant 'dot product' suggests that something similar might be possible, arranging 3-component vectors into 2-tensors with an invariant 'dot product'. The specific transformation laws for $E$ and $B$ are especially suggestive, which is why I referred to them. $\endgroup$
    – Danu
    Commented Mar 12, 2014 at 14:48
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    $\begingroup$ @user129412 Indeed, it is. Any object that can be written in the form $A=t_\mu x^\mu b_{\lambda\kappa\rho}c^\kappa f^{\lambda\rho}\dots$ (i.e. a 'full contraction' of a product of tensors of arbitrary rank) is a scalar, and therefore Lorentz invariant. $\endgroup$
    – Danu
    Commented Mar 12, 2014 at 14:57
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    $\begingroup$ let us continue this discussion in chat $\endgroup$
    – Danu
    Commented Mar 12, 2014 at 15:00

2 Answers 2

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They're lorentz scalars. Every scalar is lorentz invariant.

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$F_{\mu\nu}$ is a Lorentz tensor, easy to see by $\partial_\mu A_\nu - \partial_\nu A_\mu$, which is a 2-form. Contractions of Lorentz tensors are Lorentz tensors. $\tilde{F} = \star F$ is the Hodge dual of $F$, which is also a 2-form, hence a Lorentz tensor, therefore the same applies about its contractions. By these definitions, they are also tensors in curved spacetime.

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  • $\begingroup$ Of course, if you want to speak differential forms, then $F_{\mu\nu}F^{\mu\nu} = \star (F\wedge \star F)$ and $\tilde{F}_{\mu\nu} F^{\mu\nu} = \star (F \wedge F)$. The translation between forms and index notation is not always as obvious as one would like. :) $\endgroup$
    – Robin Ekman
    Commented Apr 12, 2014 at 18:07
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    $\begingroup$ @RobinEkman Yes, that's why I'm used to combining the two, depending on which leads to simpler expressions. For example, I've never done contractions using the Hodge dual and I prefer to replace it with the expression in terms of tensors. $\endgroup$
    – auxsvr
    Commented Apr 12, 2014 at 18:44
  • $\begingroup$ Yes, they both have their merits, a physicist worth his or her salt should be comfortable with both. $\endgroup$
    – Robin Ekman
    Commented Apr 12, 2014 at 18:59

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